9.6 Power

Work is a transfer of energy between the environment and a system. In many situations we would like to know how quickly the energy is transferred. Does the force act quickly and transfer the energy very rapidly, or is it a slow and lazy transfer of energy? If you need to buy a motor to lift 1000 kg of bricks up 20 m, it makes a big difference whether the motor has to do this in 30 s or 30 min!

The question How quickly? implies that we are talking about a rate. For example, the velocity of an object—how quickly it is moving—is the rate of change of position. So when we raise the issue of how quickly the energy is transferred, we are talking about the rate of transfer of energy. The rate at which energy is transferred or transformed is called the power P, and it is defined as

P = \frac{d E_{sys}}{d t}

(9.28)

The unit of power is the watt, which is defined as $1 watt = 1 W = 1 J/s$ . Common prefixes used with power are mW (milliwatts), kW (kilowatts), and MW (megawatts).

For example, the rope in Example 9.10 pulled with a tension of 30 N and, by doing work, transferred 90 J of energy to the system. If it took 10 s to drag the crate 3.0 m, then energy was being transferred at the rate of 9 J/s. We would say that whatever was supplying this energy—whether a human or a motor—has a “power output” of 9 W.

The idea of power as a rate of energy transfer applies no matter what the form of energy. FIGURE 9.25 shows three examples of the idea of power. For now, we want to focus primarily on work as the source of energy transfer. Within this more limited scope, power is simply the rate of doing work: $P = dW/dt$ . If a particle moves through a small displacement $d \vec{r}$ while acted on by force $\vec{F}$ , the force does a small amount of work dW given by

d W = \vec{F} \cdot d \vec{r}

Dividing both sides by dt, to give a rate of change, yields

\frac{d W}{d t} = \vec{F} \cdot \frac{d \vec{r}}{d t}

But $d \vec{r} / d t$ is the velocity $\vec{v}$ , so we can write the power as

P = \vec{F} \cdot \vec{v} = F v cos θ

(9.29)

In other words, the power delivered to a particle by a force acting on it is the dot product of the force and the particle’s velocity. These ideas will become clearer with some examples.

The figure illustrates a light bulb of 100-Watt power. Electric energy converts to light and thermal energy at 100 Joules per second.

The figure illustrates an athlete running. The chemical energy of glucose and fat is converted to mechanical energy at 350 joules per second which approximately equals half horsepower.

EXAMPLE 9.11 Lifting a motor

Suppose you need to lift a 35 kg motor onto a flatbed truck. To do so, you use a rope and a pulley attached to the ceiling to lift the motor 3.0 m in 8.0 s. What is your power output?

SOLVE The force with which you pull down on the rope and the rope’s tension pulling up on you are an action/reaction pair, so the magnitude of your pulling force $\vec{F}_{pull}$ equals the rope tension T. Thus, if we assume a frictionless pulley, the work you do to pull the rope down is exactly the same as the work that tension does to pull the motor up. If you lift the motor at a steady speed, then the rope tension is simply the weight $m g$ of the motor. The work to lift the motor is

W = T Δ y = m g Δ y = (35 kg) (9.8 {m/s}^{2}) (3.0 m) = 1030 J

Power is the rate of doing work, so

P = \frac{W}{Δ t} = \frac{1030 J}{8.0 s} = 130 J/s = 130 W

Alternatively, the tension $T = m g = 343 N$ lifts the motor at a speed of $v = Δ y / Δ t = 0.375 m / s$ , so the required power is

P = T v = (343 N) (0.375 m/s) = 130 W

EXAMPLE 9.12 Towing a car

A 1500 kg car has a front profile that is 1.6 m wide by 1.4 m high. The car’s drag coefficient is 0.50, and its coefficient of rolling friction is 0.02. What power must a tow-truck engine provide to tow this car on a level road at a steady $25 m / s$ if the drive train efficiency—the fraction of the engine’s energy output that reaches the wheels—is 90%?

SOLVE The net force on a car being pulled at steady speed is zero, so the tension force pulling the car must balance the retarding friction and drag forces. Tension does work by pulling the car through a displacement, and the required power—given by Equation 9.29 with $F = T$ and $θ = 0 °$ —is $P = T v$ . In this case, the work does not increase the car’s kinetic energy but is dissipated as thermal energy by friction and drag.

The tension force that balances the two opposing forces is

T = f_{r} + F_{drag}

We can use the results of Chapter 6, where both rolling friction and drag were introduced, to compute

T = μ_{r} m g + \frac{1}{2} C_{d} ρ A v^{2} = 294 N + 420 N = 714 N

Here $A = (1.6 m) \times (1.4 m)$ is the front cross-section area of the car, and we used $1 {.2 kg/m}^{3}$ as the density of $20 ° C$ air. Thus the power required to pull the car at $25 m / s$ is

P = T v = (714 N) (25 m/s) = 18, 000 W = 24 hp

This is the required pulling power, but the tow-truck engine’s power output must be larger to compensate for friction losses in the drivetrain. For a 90% drivetrain efficiency—a typical value—the required engine power is

P_{engine} = \frac{18, 000 W}{0.90} = 20, 000 W = 27 hp

REVIEW 27 hp is the additional engine power needed to tow the car; that is, in addition to the power needed to drive the tow truck at a speed of $25 m/s$ . A truck driving at this speed uses roughly 50 hp, so towing increases the engine’s power output by roughly 50%. A tow-truck engine might be rated at 400 hp, but most of that power is reserved for fast acceleration and climbing hills.

CHAPTER 9 CHALLENGE EXAMPLE Stopping a brick

A 25.0-cm-long spring stands vertically on the ground, with its lower end secured in a base. A 1.5 kg brick is held 40 cm directly above the spring and dropped onto the spring. The spring compresses to a length of 17.0 cm before starting to launch the brick back upward. What is the spring’s spring constant?

MODEL Let the system consist of only the brick, which is modeled as a particle. Assume that the spring is ideal. Both gravity and, after contact, the spring do work on the brick.

VISUALIZE FIGURE 9.26 is a before-and-after pictorial representation. We’ve chosen to place the origin of the y-axis at the equilibrium position of the spring’s upper end. The length of the spring is not relevant, but the difference between its before and after lengths—8.0 cm—is the spring’s maximum compression. The point of maximum compression is a turning point for the brick, as it reverses direction and starts moving back up, so the brick’s instantaneous velocity is zero.

The figure illustrates the before and after pictorial representation of the collision of brick on a vertical spring of spring constant k.

FIGURE 9.26 Full Alternative Text

SOLVE At first, this seems like a two-part problem: free fall until hitting the spring, then deceleration as the spring compresses. But by using the energy principle, $Δ E_{sys} = Δ K = W_{tot} = W_{G} + W_{Sp}$ , we can do it in one step. Interestingly, $Δ K = 0$ because the brick is instantaneously at rest at the beginning and again at the point of maximum spring compression. Consequently, $W_{tot} = 0$ . That’s not a difficulty because gravity does positive work (the downward gravitational force is in the direction of the brick’s displacement) while the spring does negative work (the upward spring force is opposite the displacement).

The work done by gravity is

W_{G} = (F_{G})_{y} Δ y_{brick} = - m g Δ y_{brick}

where $Δ y_{brick}$ is the brick’s total displacement. The negative sign comes from $(F_{G})_{y} = - m g$ , but $W_{G}$ is positive because $Δ y_{brick} = y_{1} - y_{0} = - 0.48 m$ is also negative. It may seem strange that calculating the work done by gravity is so simple when the brick first accelerates, then slows quickly after hitting the spring. But work depends on only the displacement, not how fast or slow the object is moving.

We have to be careful with the spring because its displacement is not the same as the brick’s displacement. The spring begins compressing only when contact is made, so $Δ y_{Sp 0} = 0 m$ at the instant the brick is released. The work done by the spring then continues until maximum compression, when the spring’s displacement is $Δ y_{Sp 1} = - 0.080 m$ . The work done by the spring, Equation 9.25, is thus

W_{Sp} = - (\frac{1}{2} k {(Δ y_{Sp 1})}^{2} - \frac{1}{2} k {(Δ y_{Sp 0})}^{2}) = - \frac{1}{2} k {(Δ y_{Sp 1})}^{2}

With this information about the two works, the energy principle is

Δ K = 0 = W_{G} + W_{Sp} = - m g Δ y_{brick} - \frac{1}{2} k (Δ y_{Sp 1})^{2}

Solving for the spring constant gives

\begin{array}{l} k & = - \frac{2 m g Δ y_{brick}}{{(Δ y_{Sp 1})}^{2}} = - \frac{2 (1.5 kg)(9.80 {m/s}^{2}) (- 0.48 m)}{{(- 0.080 m)}^{2}} \\ = 2200 N/m \end{array}

REVIEW 2200 N/m is a fairly large spring constant, but that’s to be expected for a spring that’s going to stop a falling $\approx 3$ pound brick. The complexity of this problem was not the math, which was fairly simple, but the reasoning. It’s a good illustration of how to apply energy reasoning to other problems.

9.6 Power

FIGURE 9.25 Examples of power.

EXAMPLE 9.11 Lifting a motor

EXAMPLE 9.12 Towing a car

CHAPTER 9 CHALLENGE EXAMPLE Stopping a brick

FIGURE 9.26 Pictorial representation of the brick and spring.